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The number of electrical outages in a city varies from day to day. Assume that the number of electrical outages ( x ) in the cit

Mrac [35]

Answer:

The mean is 0.26

The standard deviation is 0.5765

Step-by-step explanation:

An image showing the step by step solution is attached.

5 0

3 years ago

Read 2 more answers

10|x+8| - 13 = -2|x+8|+11

PilotLPTM [1.2K]

Use the app Socratic

4 0

3 years ago

uppose a large telephone manufacturer has a problem with excessive customer complaints and consequent returns of the phones for

vodka [1.7K]

Answer:

The answer is "294.2075".

Step-by-step explanation:

Given:

$\hat{p} = 0.5 \\\\1 - \hat{p} = 1 - 0.5 = 0.5\\\\E = 4\% = 0.04\\\\$

At $83\%$ confidence level the z is ,

$\alpha = 1 - 83\% = 1 - 0.83 = 0.17\\\\\frac{\alpha}{2} = \frac{0.17}{2} = 0.085\\\\Z_{\frac{\alpha}{2}} = Z_{0.085} = 1.3722\\\\n = (\frac{Z_{\frac{\alpha}{2}}}{E})^2 \times \hat{p} \times (1 - \hat{p})\\\\$

$= (\frac{1.3722}{0.04})^2 \times 0.5 \times 0.5\\\\= (34.305)^2 \times 0.5 \times 0.5\\\\= 1,176.83 \times 0.5 \times 0.5\\\\= 1,176.83 \times 0.25\\\\=294.2075$

5 0

3 years ago

Consider testing H^0: u=20 against H^a: u<20 where u is the mean number of latex gloves used per week by all hospital employe

iragen [17]

Answer:

Part a: P-value

The first step is calculate the degrees of freedom, on this case:

$df=n-1=46-1=45$

Since is a one sided test the p value would be:

$p_v =P(t_{(45)}$

Part b: Conclusion

If we compare the p value and the significance level given $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to fail reject the null hypothesis, and we can conclude that the true mean is not lower than 20 at 1% of signficance.

a. There is insufficient evidence to reject h^0

Step-by-step explanation:

Data given and notation

$\bar X=19.1$ represent the sample mean

$s=11.8$ represent the sample standard deviation

$n=46$ sample size

$\mu_o =20$ represent the value that we want to test

$\alpha=0.01$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is lower than 20, the system of hypothesis would be:

Null hypothesis: $\mu \geq 20$

Alternative hypothesis: $\mu < 20$

If we analyze the size for the sample is > 30 but we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{19.1-20}{\frac{11.8}{\sqrt{46}}}=-0.517$

Part a: P-value

The first step is calculate the degrees of freedom, on this case:

$df=n-1=46-1=45$

Since is a one sided test the p value would be:

$p_v =P(t_{(45)}$

Part b: Conclusion

If we compare the p value and the significance level given $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to fail reject the null hypothesis, and we can conclude that the true mean is not lower than 20 at 1% of signficance.

a. There is insufficient evidence to reject h^0

5 0

3 years ago

A farmer has a crop of 2,000 orange trees at the beginning of this year. Each year, the farmer plans to plant 250 new trees. Ass

lawyer [7]

Answer:

y=250x+2000

Step-by-step explanation:

The farmer starts with 2000 trees and adds 250 each year.

Year 0 2000

Year 1 2000 + 250 = 2000+ 1(250)

Year 2 2000 + 250 +250 = 2000 + 2(250)

Year 3 2000 + 250 + 250 + 250 = 2000 + 3(250)

Year 4 2000 + 250 + 250 + 250 + 250 = 2000 + 4 (250)

.....

Year X 2000 + 250x

5 0

3 years ago